Ritesh travels at $1 ext{ kmph}$ to reach a station $500 ext{ m}$ away from his house to catch a train. He starts walking but after $4$ minutes, he realizes he forgot a document at home and returns at the same speed. What should be his speed (in kmph) on the way back to catch the train on time?

Question

Ritesh travels at $1 	ext{ kmph}$ to reach a station $500 	ext{ m}$ away from his house to catch a train. He starts walking but after $4$ minutes, he realizes he forgot a document at home and returns at the same speed. What should be his speed (in kmph) on the way back to catch the train on time?

Accepted Answer

**Solution:**

1. Convert all units to consistent terms:
   - Distance to station = $500 	ext{ m} = 0.5 	ext{ km}$
   - Speed initially = $1 	ext{ kmph}$
   - Time before returning = $4$ minutes = $\frac{4}{60} = \frac{1}{15}$ hours

2. Calculate the distance Ritesh covers before turning back:
   $$	ext{Distance covered} = 	ext{Speed} 	imes 	ext{Time} = 1 	imes \frac{1}{15} = \frac{1}{15} 	ext{ km}$$

3. Distance remaining to station after returning home:
   - He returns the same distance $\frac{1}{15} 	ext{ km}$ back to home.
   - Total distance to station is $0.5 	ext{ km}$.

4. Total time to reach station if he had gone directly:
   $$	ext{Time} = \frac{0.5}{1} = 0.5 	ext{ hours}$$

5. Time spent so far:
   - Going forward: $\frac{1}{15}$ hours
   - Returning back: $\frac{1}{15}$ hours
   - Total spent: $\frac{2}{15}$ hours

6. Remaining time to reach station on second attempt:
   $$0.5 - \frac{2}{15} = \frac{15}{30} - \frac{4}{30} = \frac{11}{30} 	ext{ hours}$$

7. Distance to cover now:
   $$0.5 	ext{ km}$$

8. Required speed to cover $0.5 	ext{ km}$ in $\frac{11}{30}$ hours:
   $$	ext{Speed} = \frac{	ext{Distance}}{	ext{Time}} = \frac{0.5}{\frac{11}{30}} = 0.5 	imes \frac{30}{11} = \frac{15}{11} \approx 1.36 	ext{ kmph}$$

**Answer:** $1.36 	ext{ kmph}$

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